Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $z = \dfrac{x^2 - 4x - 21}{5x - 5} \div \dfrac{10x + 30}{-6x + 6} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{x^2 - 4x - 21}{5x - 5} \times \dfrac{-6x + 6}{10x + 30} $ First factor the quadratic. $z = \dfrac{(x + 3)(x - 7)}{5x - 5} \times \dfrac{-6x + 6}{10x + 30} $ Then factor out any other terms. $z = \dfrac{(x + 3)(x - 7)}{5(x - 1)} \times \dfrac{-6(x - 1)}{10(x + 3)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (x + 3)(x - 7) \times -6(x - 1) } { 5(x - 1) \times 10(x + 3) } $ $z = \dfrac{ -6(x + 3)(x - 7)(x - 1)}{ 50(x - 1)(x + 3)} $ Notice that $(x - 1)$ and $(x + 3)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ -6\cancel{(x + 3)}(x - 7)(x - 1)}{ 50(x - 1)\cancel{(x + 3)}} $ We are dividing by $x + 3$ , so $x + 3 \neq 0$ Therefore, $x \neq -3$ $z = \dfrac{ -6\cancel{(x + 3)}(x - 7)\cancel{(x - 1)}}{ 50\cancel{(x - 1)}\cancel{(x + 3)}} $ We are dividing by $x - 1$ , so $x - 1 \neq 0$ Therefore, $x \neq 1$ $z = \dfrac{-6(x - 7)}{50} $ $z = \dfrac{-3(x - 7)}{25} ; \space x \neq -3 ; \space x \neq 1 $